# Spivak's proof of Inverse Function Theorem

I am having trouble with Spivak's proof of the Inverse Function Theorem in his Calculus on Manifolds:

2-11 Theorem (Inverse Function Theorem). Suppose that $f: \mathbb{R}^n\to\mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and det $f'(a)\neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f:V\to W$ has a continuous inverse $f^{-1}:W\to V$ which is differentiable and for all $y\in W$ satisfies $$(f^{-1})'(y) = [f'(f^{-1}(y))]^{-1}$$ Proof. Let $\lambda$ be the linear transformation $Df(a)$. Then $\lambda$ is non-singular, since det $f'(a)\neq 0$. Now $D(\lambda^{-1}\circ f)(a) = D(\lambda^{-1})(f(a))\circ Df(a) = \lambda^{-1}\circ Df(a)$ is the identity linear transformation. If the theorem is true for $\lambda^{-1}\circ f$, it is clearly true for $f$...

How is the theorem true for $f$ if it is true for $\lambda^{-1}\circ f$?

• I have not read Spivak's book, but I would have guessed $\lambda$ is the constant map $x\rightarrow \lambda x$, which does not influence $f$'s differentiablity since $\lambda f$ is just multiplying $f$ by a constant. Sep 1, 2012 at 20:35
• I think the key difficulty around this point isn't that $f$ is differentiable if $\lambda$$f$ is but that the formula for the inverse remains true. May 27 at 8:50

$$\lambda\colon \mathbb{R}^n\to\mathbb{R}^n$$ is a bijection and $$\lambda$$ and $$\lambda^{-1}$$ are both continuously differentiable. Note that $$\lambda'(z) = \lambda$$ for all $$z \in \mathbb{R}$$.

Let $$g = \lambda^{-1}\circ f$$. Suppose the theorem is true for $$g$$. Then there is an open set $$V'$$ containing $$a$$ and an open set $$W'$$ containing $$g(a)$$ such that $$g:V'\to W'$$ has a continuous inverse $$g^{-1}:W'\to V'$$ which is differentiable and for all $$y\in W'$$ satisfies $$(g^{-1})'(y) = [g'(g^{-1}(y))]^{-1}$$ Then $$\lambda(W')$$ is open and $$f = \lambda\circ g:V'\to \lambda(W')$$ has a continuous inverse $$g^{-1}\circ \lambda^{-1}:\lambda(W')\to V'$$.

By the chain rule, for all $$z \in \lambda(W')$$, $$(f^{-1})'(z) = (g^{-1}\circ\lambda^{-1})'(z) = (g^{-1})'(\lambda^{-1}(z))\circ \lambda^{-1} = [g'(g^{-1}(\lambda^{-1}(z)))]^{-1}\circ \lambda^{-1} = [g'(f^{-1}(z))]^{-1}\circ \lambda^{-1} = [\lambda\circ g'(f^{-1}(z)]^{-1} = [f'(f^{-1}(z))]^{-1}$$

• Why is $\lambda$ a bijection? May 27, 2018 at 12:07
• $\lambda$ is non-singular (since $\det f'(a) \not= 0$). Hence it estabilish a one-to-one correspondence between the function domain and image (i.e. it is a bijection) Mar 28, 2019 at 7:53
• $Df(a)$ is a linear mapping, but $f'(a)$ is a matrix. So, $\lambda^{'}(z)$ is a matrix but $\lambda$ is a linear mapping. So, $\lambda^{'}(z)\neq\lambda$. Aug 27, 2022 at 20:08

Suppose the theorem holds for all functions $f$ such that $Df(a) = I$. What Spivak has shown is that for $\lambda = Df(a)$, $D(\lambda^{-1} \circ f(a)) = I$. By the assumption above, the theorem must be true for $\lambda^{-1}\circ f$. That means there exists $g$ which is the continuously differentiable inverse of $\lambda^{-1} \circ f$. This implies that $(g \circ \lambda^{-1}) \circ f = I$. Now $g \circ \lambda^{-1}$ is continuously differentiable, and so the theorem is true for $f$.

• I dont understand the last statement. How does $g \circ \lambda ^{-1}$being continuously differentiable imply the theorem is true for $f$? May 27, 2018 at 12:11
• Notice how he placed the parentheses around $g\circ\lambda^{-1}$. This has no effect on computing the compositions; he simply did it to show you that $g\circ\lambda^{-1}$ is in fact the inverse of $f$ that we are looking for! And in addition, this inverse is continuously differentiable since it is just the map $g$ and we generated $g$ as a result of applying the Inverse function Theorem to $\lambda^{-1} \circ f(a)$, so $g$ certainly satisfies all the properties that we'd want from our inverse. Jun 7, 2018 at 0:01
• The inverse is continuous and differentiable and may not be continuously differentiable Sep 23, 2020 at 19:39

Let $$f: \mathbb{R}^n\to\mathbb{R}^n$$ be a continuously differentiable function in an open set containing $$a$$, and det $$f'(a)\neq 0$$.
Let $$\lambda := Df(a)$$.
Then $$\lambda$$ is a non-singular linear transformation because $$\det f'(a) \neq 0$$.
Let $$g := \lambda^{-1} \circ f$$ and $$f = (f^1, \cdots, f^n)$$ and $$g = (g^1, \cdots, g^n)$$.
Then we can write $$g^i(x)$$ as $$g^i(x) = \mu_{i 1} f^1(x) + \cdots + \mu_{i n} f^n(x)$$ for some real numbers $$\mu_{i 1}, \cdots, \mu_{i n}$$.
Then each $$g^i$$ is a continuously differentiable function in an open set containing $$a$$ because each $$f^i$$ is a continuously differentiable function in an open set containing $$a$$.
So, $$g:\mathbb{R}^n\to\mathbb{R}^n$$ is continuously differentiable in an open set containing $$a$$.

$$D(\lambda^{-1} \circ f)(a) = D\lambda^{-1}(f(a)) \circ Df(a) = \lambda^{-1} \circ Df(a) = I$$.
So, $$g'(a) = I_n$$.
So, $$\det g'(a)=1\neq 0$$.

Then by the Spivak's proof in the text, there is an open set $$V$$ containing $$a$$ and an open set $$W$$ containing $$g(a)$$ such that $$g:V\to W$$ has a continuous inverse $$g^{-1} : W \to V$$ which is differentiable and for all $$y\in W$$ satisfies $$(g^{-1})^{'}(y)=[g^{'}(f^{-1}(y))]^{-1}.$$

$$f = \lambda \circ g$$.

The image of $$\lambda : W \to \mathbb{R}^n$$ is $$\lambda (W)$$.
$$\lambda : W \to \lambda (W)$$ is injective because $$\lambda : \mathbb{R}^n \to \mathbb{R}^n$$ is non-singular.
So, $$\lambda : W \to \lambda (W)$$ is bijective.
So $$f=\lambda\circ g : V \to \lambda (W)$$ is bijective.

$$\lambda W$$ is an open set. A proof of this fact is the following:
Let $$\lambda(w) \in \lambda (W)$$.
By the problem 1-10 on p.5, there exists a positive real number $$M$$ such that $$|\lambda^{-1} (h)| \leq M |h|$$ for any $$h \in \mathbb{R}^n$$.
Because $$W$$ is open, there exists $$\epsilon > 0$$ such that for any $$w' \in \mathbb{R}^n$$, if $$|w' - w| < \epsilon$$, then $$w' \in W$$.
Let $$u'$$ be an arbitrary element of $$\mathbb{R}^n$$ such that $$|u' - \lambda w| < \frac{\epsilon}{M}$$.
Then, $$|\lambda^{-1} (u') - w| = |\lambda^{-1} (u') - \lambda^{-1} (\lambda w)| = |\lambda^{-1} (u' - \lambda w)| \leq M |u' - \lambda w| < M \frac{\epsilon}{M} = \epsilon$$.
So, $$\lambda^{-1} (u') \in W$$.
So, $$u' = \lambda(\lambda^{-1}(u')) \in \lambda (W)$$.
So, $$\lambda (W)$$ is open.

Now $$f : V \to \lambda (W)$$ is a continuously differentiable bijective function from an open set $$V$$ to an open set $$\lambda (W)$$.

$$f^{-1} = g^{-1} \circ \lambda^{-1}$$.
$$g^{-1}$$ is differentiable on $$W$$ and $$\lambda^{-1}$$ is differentiable on $$\lambda (W)$$.
So, $$f^{-1}$$ is differentiable on $$\lambda (W)$$ by Chain Rule.
Let $$\lambda^{'}(x)=A$$.
Then, \begin{align*} (f^{-1})^{'}(y) &= (g^{-1}\circ\lambda^{-1})^{'}(y) = (g^{-1})^{'}(\lambda^{-1}(y))\cdot(\lambda^{-1})^{'}(y)\\ &=[g^{'}(g^{-1}(\lambda^{-1}(y)))]^{-1}\cdot A^{-1} =[g^{'}(f^{-1}(y))]^{-1}\cdot A^{-1}\\ &=[A\cdot g^{'}(f^{-1}(y))]^{-1} =[(\lambda\circ g)^{'}(f^{-1}(y))]^{-1}\\ &=[f^{'}(f^{-1}(y))]^{-1}. \end{align*}

• I totally agree that this discussion is needed. I've googled a few of these proofs and there is far too little justification of the assertion that we can assume WLOG that the Jacobian is the identity. May 27 at 8:45